First we will find the speed of the ball just before it will hit the floor
so in order to find the speed of the cart we will first use energy conservation



So by solving above equation we will have

now in order to find the momentum we can use




Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
Answer:
v= 2413.5 m/s
Explanation:
maximum change of speed of rocket
=(initial exhaust velocity)×ln [(initialmass/finalmass)]
let initial mass= m
final mass = m-m(4/5) = m/5
[since the 80% of mass which is fuel is exhausted]
V-0 = 1500 ln (1/0.2)
V= 1500×1.609 = 2413.5 m/s
therefore, its exhaust speed v= 2413.5 m/s
Answer:
Explanation:
In first case we are interested in one time 6 in six rolls
Thus probability = number of chances required/Total chances
= 1/6
Similarly in the second case probability = 2/12 = 1/6
In the same way in last case probability = 100/600 = 1/6
The probability is the same . Thus all the cases has equal chances
Answer:
The energy of this particle in the ground state is E₁=1.5 eV.
Explanation:
The energy
of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

So we can rewrite the energy in the ground state as:



Finally
